7.4  Singularities, Zeros, and Poles

    Recall that the point is called a singular point, or  singularity of the complex function f(z) if  f  is not analytic at  ,  but every neighborhood    contains at least one point at which f(z) is analytic.  For example, the function    is not analytic at  ,  but is analytic for all other values of z.  Thus the point    is a singular point of f(z).  As another example,  consider  .  We saw in Section 5.2 that g(z) is  analytic for all z except at the origin and at all points on the negative real-axis.  Thus, the origin and each point on the negative real axis is a singularity of  .

    The point is called a isolated singularity of the complex function f(z) if  f  is not analytic at  ,  but there exists a real number    such that f(z) is analytic everywhere in the punctured disk  .  The function    has an isolated singularity at  .  

    The function  ,  however, the singularity at    (or at any point of the negative real axis) that is not isolated, because any neighborhood of  contains points on the negative real axis, and    is not analytic at those points.  Functions with isolated singularities have a Laurent series because the punctured disk     is the same as the annulus  .  The logarithm function    does not have a Laurent series at any point    on the negative real-axis.  We now look at this special case of Laurent's theorem in order to classify three types of isolated singularities.

Definition 7.5 (Removable Singularity, Pole of order k, Essential Singularity). Let f(z) have an isolated singularity at    with Laurent series expansion  

                valid for    .  

Then we distinguish the following types of singularities at  .  

(i)      If  ,  then we say that f(z) has a removable singularity at  .  

(ii)      If k is a positive integer such that    but ,  then we say that f(z) has a pole of order k at .  

(iii)      If for infinitely many negative integers n, then we say that f(z) has an essential singularity at  .  

    Let's investigate some examples of these three cases.

(i).  If f(z) has a removable singularity at , then it has a Laurent series  

            valid for    .  

    Theorem 4.17 implies that the power series for f(z) defines an analytic function in the disk  .  
    If we use this series to define  ,  then the function f(z) becomes analytic at , removing the singularity.  

    For example, consider the function  .  It is undefined at and has an isolated singularity at , as the Laurent series for f(z) is  

              

valid for .  

We can remove this singularity if we define  ,  for then f(z) will be analytic at in accordance with Theorem 4.17.  

Exploration 1.

    Another example is  ,  which has an isolated singularity at the point , as the Laurent series for g(z) is

            

valid for .  If we define  ,  then g(z) will be analytic for all z.

Exploration 2.

(ii).  If f(z) has a pole of order k at , the Laurent series for f(z) is

            valid for    .  

where  .  

Extra Example 1. The following example will help this concept.  Consider the function  .  The leading term in the Laurent series expansion  S(z)  is    and  S(z)  goes to in the same manner as  .            

Explore Solution Extra Example 1.

    Another example is;  

              

has a pole of order    at  .

Exploration 3.

    If f(z) has a pole of order 1 at , we say that f(z) has a simple pole at .   

    For example,  

              

has a simple pole at  .  

Exploration 4.

(iii).  If infinitely many negative powers of occur in the Laurent series, then f(z) has an essential singularity at .  For example,  

              

has an essential singularity at the origin.  

Exploration 5.

Definition 7.6 (Zero of order k).  A function f(z) analytic in    has a zero of order k at the point    if and only if  

            ,  and  .

A zero of order one is sometimes called a simple zero.  

Theorem 7.10.  A function analytic in has a zero of order k at the point   iff its Taylor series given by has

            .  

Proof.

Example 7.10.  From Theorem 7.10 we see that the function

               

has a zero of order at .  Definition 7.6 confirms this fact because   

              

Then,  ,  but  .  

Explore Solution 7.10.

Theorem 7.11.  Suppose f(z) is analytic in .  Then f(z) has a zero of order k at the point if and only if it can be expressed in the form

(7-35)            ,

where g(z) is analytic at .  

Proof.

    An immediate consequence of Theorem 7.11 is Corollary 7.4.  The proof is left as an exercise.

Corollary 7.4.  If f(z) and g(z) are analytic at  ,  and have zeros of orders  m  and  n,  respectively at  ,  then their product    has a zero of order  .  

Proof.

Example 7.11.  Let .  Then f(z) can be factored as the product of    and  ,  which have zeros of orders and , respectively, at .  
Hence   is a zero of order 4 of  f(z).

Explore Solution 7.11.

    Theorem 7.12 gives a useful way to characterize a pole.  

Theorem 7.12.  A function f(z) analytic in the punctured disk has a pole of order k at if and only if it can be expressed in the form
            
(7-37)            ,

where the function h(z) is analytic at the point .  

Proof.

    Corollaries 7.5-7.8 are useful in determining the order of a zero or a pole.  The proofs follow easily from Theorems 7.10 and 7.12 and are left as exercises.

Corollary 7.5.  If f(z) is analytic and has a zero of order k at the point  ,  then    has a pole of order k at  .  

Proof.

Corollary 7.6. If  f(z) has a pole of order k at the point , then has a removable singularity at .  If we define ,  then g(z)  has a zero of order k at .  

Proof.

Corollary 7.7.  If f(z) and g(z) have poles of orders  m  and  n,  respectively at the point  , then their product    has a pole of order  .  

Proof.

Corollary 7.8.  Let f(z) and g(z) be analytic with  zeros of orders  m  and  n,  respectively at  .  Then their quotient   has the following behavior:

(i)  If  ,  then h(z) has a removable singularity at  .   If we define  ,  then h(z) has a zero of order  .

(ii)  If  ,  then h(z) has a pole of order  .

(iii)  If  ,  then h(z) has a removable singularity  at  ,  and can be defined so that h(z) is analytic at  ,  by  .  

Proof.

Example 7.12.  Locate the zeros and poles of  ,  and determine their order.

Solution.  In Section 5.4 we saw that the zeros of    occur at the points  ,  where n is an integer.  Because  ,  the zeros of f(z) are simple.  Similarly, the function    has simple zeros at the points and  ,  where n is an integer.  From the information given, we find that    behaves as follows:

         i.    h(z)  has simple zeros at  ,  where  ;

         ii.   h(z)  has simple poles at  ,  where n is an integer;  and

         iii.  h(z)  is analytic at if we define  .

Explore Solution 7.12.

Example 7.13.  Locate the poles of , and specify their order.

Solution.  The roots of the quadratic equation    occur at the points .  If we replace z with in this equation, the function    has simple zeros at the points  .  Corollary 7.5 implies that g(z)  has simple poles at  .

Explore Solution 7.13.

Example 7.14.  Locate the zeros and poles of , and determine their order.

Solution. The function    has a zero of order    at    and simple zeros at the points  .  Corollary 7.5 implies that g(z) has a pole of order 3 at the point    and simple poles at the points  .  

Explore Solution 7.14.

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(c) 2006 John H. Mathews, Russell W. Howell